Problem: $g(n)=25-49(n-1)$ Complete the recursive formula of $g(n)$. $g(1)=$
Answer: From the explicit formula, ${25}{-49}(n-1)$, we can tell that the first term of the sequence is ${25}$ and the common difference is ${-49}$. This is the recursive formula of the sequence: $\begin{cases} g(1)={25} \\\\ g(n)=g(n-1)+({-49}) \end{cases}$